What is the integral from 0 to 4 of sqrt(9t^2+t^4) ?

The integral from 0 to 4 of sqrt(9t^2+t^4) is 98/3

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integral from 0 to 4 of sqrt(9t^2+t^4)

\int_{0}^{4} 9 t^{2} + t^{4} d t

\frac{98}{3}

Decimal

32.66666 \dots

Solution steps

\int_{0}^{4} 9 t^{2} + t^{4} d t

Multiply by the conjugate of

9 t^{2} + t^{4} : \frac{81 t ^{4} - t ^{8}}{9 t ^{2} - t ^{4}}

= \int_{0}^{4} \frac{81 t ^{4} - t ^{8}}{9 t ^{2} - t ^{4}} d t

Find undefined (singularity) points:

t = 3

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{3} \frac{81 t ^{4} - t ^{8}}{9 t ^{2} - t ^{4}} d t + \int_{3}^{4} \frac{81 t ^{4} - t ^{8}}{9 t ^{2} - t ^{4}} d t

\int_{0}^{3} \frac{81 t ^{4} - t ^{8}}{9 t ^{2} - t ^{4}} d t = 182 - 9

\int_{3}^{4} \frac{81 t ^{4} - t ^{8}}{9 t ^{2} - t ^{4}} d t = \frac{125}{3} - 182

= 182 - 9 + \frac{125}{3} - 182

Simplify

182 - 9 + \frac{125}{3} - 182 : \frac{98}{3}

= \frac{98}{3}

What is the integral from 0 to 4 of sqrt(9t^2+t^4) ?
The integral from 0 to 4 of sqrt(9t^2+t^4) is 98/3